r - 一阶差分数据框

Question

我有以下数据框:

>dados

COUNTRY   Year   CO2 emissions Pop. Growth(%)
Argentina  1994      1.23         0.3
Argentina  1995      1.26         0.2
Argentina  1996      1.28         0.4
Argentina  1997      1.24         0.2
Brazil     1994      1.54         0.7
Brazil     1995      1.59         0.6
Brazil     1996      1.60         0.9
Brazil     1997      1.58         1.3

我想首先区分变量 CO2 emissions和 Pop. Growth(%)对于每个国家。我已经尝试过函数 dados[,2:4] <- diff(dados[,2:4])但它返回了错误:

"Error in r[i1] - r[-length(r):-(length(r) - lag + 1L)] : non-numeric argument to binary operator"

Answer 1

这是 dplyr:

library(dplyr)

df %>%
  group_by(COUNTRY) %>%
  mutate_at(vars(CO2_emissions:Pop_Growth), funs(.-lag(.)))

编辑:从 dplyr 0.8.0 开始，funs() 已软弃用。对较新版本的 dplyr

使用以下代码

df %>%
  group_by(COUNTRY) %>%
  mutate_at(vars(CO2_emissions:Pop_Growth), list(~ .x - lag(.x)))

输出:

# A tibble: 8 x 4
# Groups:   COUNTRY [2]
  COUNTRY    Year CO2_emissions Pop_Growth
  <fct>     <int>         <dbl>      <dbl>
1 Argentina  1994         NA        NA    
2 Argentina  1995          0.03     -0.100
3 Argentina  1996          0.02      0.2  
4 Argentina  1997         -0.04     -0.2  
5 Brazil     1994         NA        NA    
6 Brazil     1995          0.05     -0.100
7 Brazil     1996          0.01      0.3  
8 Brazil     1997         -0.02      0.4 

数据:

df = structure(list(COUNTRY = structure(c(1L, 1L, 1L, 1L, 2L, 2L, 
2L, 2L), .Label = c("Argentina", "Brazil"), class = "factor"), 
    Year = c(1994L, 1995L, 1996L, 1997L, 1994L, 1995L, 1996L, 
    1997L), CO2_emissions = c(1.23, 1.26, 1.28, 1.24, 1.54, 1.59, 
    1.6, 1.58), Pop_Growth = c(0.3, 0.2, 0.4, 0.2, 0.7, 0.6, 
    0.9, 1.3)), .Names = c("COUNTRY", "Year", "CO2_emissions", 
"Pop_Growth"), class = "data.frame", row.names = c(NA, -8L))